Numerical Solutions for Engineering Principles Assignments

This page contains step-by-step solutions for Engineering Principles assignments related to measurement and mensuration, kinematics and more. It includes solutions for questions related to distance, velocity, acceleration, time and height.

Numerical
1 | P a g e
Engineering principles

Numerical
Contents
Solution of Assignment_3 question...............................................................................................11
Solution 1a.................................................................................................................................11
Solution 1b.................................................................................................................................11
Solution 1c.................................................................................................................................11
Solution 2a.................................................................................................................................12
Solution 2b.................................................................................................................................12
Solution 2c.................................................................................................................................12
Solution 2d.................................................................................................................................12
Solution 2e.................................................................................................................................13
Solution 2f.................................................................................................................................13
Solution 2g.................................................................................................................................13
Solution 3a.................................................................................................................................14
Solution 3b.................................................................................................................................14
Solution 3c.................................................................................................................................14
Solution 4a.................................................................................................................................14
Solution 4b.................................................................................................................................14
Solution 5a.................................................................................................................................15
Solution 5b.................................................................................................................................15
Solution 6a.................................................................................................................................16
Solution 6b.................................................................................................................................16
Worksheet One – Measurement and Mensuration.........................................................................16
Solution 1a.................................................................................................................................16
Solution 1b.................................................................................................................................16
Solution 2a.................................................................................................................................16
Solution 2b.................................................................................................................................17
Solution 3...................................................................................................................................17
Solution 4...................................................................................................................................17
Solution 5...................................................................................................................................17
Solution 6...................................................................................................................................17
Solution 7...................................................................................................................................17
2 | P a g e

Numerical
Solution 8...................................................................................................................................18
Solution 9a.................................................................................................................................18
Solution 9b.................................................................................................................................18
Solution 10.................................................................................................................................18
Solution 11.................................................................................................................................18
Solution 12.................................................................................................................................19
Solution 13.................................................................................................................................19
Solution 14.................................................................................................................................19
Solution 15.................................................................................................................................19
Solution 16.................................................................................................................................19
Solution 17a...............................................................................................................................19
Solution 17b...............................................................................................................................20
Solution 17c...............................................................................................................................20
Solution 18a...............................................................................................................................20
Solution 18b...............................................................................................................................20
Solution 18c...............................................................................................................................20
Solution 19a...............................................................................................................................20
Solution 19b...............................................................................................................................20
Solution 20a...............................................................................................................................20
Solution 20b...............................................................................................................................20
Solution 21.................................................................................................................................20
Solution 22.................................................................................................................................21
Solution 23.................................................................................................................................21
Solution 24.................................................................................................................................21
Solution 24a...............................................................................................................................21
Solution 24b...............................................................................................................................21
Solution 24c...............................................................................................................................21
Solution 24d...............................................................................................................................21
Solution 25.................................................................................................................................22
Solution 26.................................................................................................................................22
Solution 27.................................................................................................................................22
3 | P a g e

Numerical
Solution 28a...............................................................................................................................22
Solution 28b...............................................................................................................................22
Worksheet Two – Kinematics.......................................................................................................23
Solution 1a.................................................................................................................................23
Solution 1b.................................................................................................................................23
Solution 1c.................................................................................................................................23
Solution 2a.................................................................................................................................23
Solution 2b.................................................................................................................................23
Solution 2c.................................................................................................................................23
Solution 3a.................................................................................................................................23
Solution 3b.................................................................................................................................23
Solution 3c.................................................................................................................................23
Solution 4a.................................................................................................................................23
Solution 4b.................................................................................................................................24
Solution 4c.................................................................................................................................24
Solution 5a.................................................................................................................................24
Solution 5b.................................................................................................................................24
Solution 6...................................................................................................................................24
Solution 6...................................................................................................................................24
Solution 7...................................................................................................................................24
Solution 8...................................................................................................................................25
Solution 9...................................................................................................................................25
Solution 10.................................................................................................................................25
Solution 11a...............................................................................................................................25
Solution 11b...............................................................................................................................25
Solution 11c...............................................................................................................................25
Solution 11d...............................................................................................................................25
Solution 12.................................................................................................................................25
Solution 13.................................................................................................................................26
Solution 14a...............................................................................................................................26
Solution 14b...............................................................................................................................26
4 | P a g e

Numerical
Solution 15a...............................................................................................................................26
Solution 15b...............................................................................................................................26
Solution 16.................................................................................................................................26
Solution 17a...............................................................................................................................26
Solution 17b...............................................................................................................................26
Solution 18a...............................................................................................................................27
Solution 18b...............................................................................................................................27
Solution 19.................................................................................................................................27
Solution 20.................................................................................................................................27
Solution 21.................................................................................................................................28
Solution 22.................................................................................................................................28
Solution 23a...............................................................................................................................28
Solution 23b...............................................................................................................................28
Solution 23b...............................................................................................................................28
Solution 24a...............................................................................................................................29
Solution 24b...............................................................................................................................29
Solution 24c...............................................................................................................................29
Solution 25a...............................................................................................................................29
Solution 25b...............................................................................................................................30
Solution 25c...............................................................................................................................30
Solution 26a...............................................................................................................................30
Solution 26b...............................................................................................................................30
Solution 26c...............................................................................................................................30
Solution 26d...............................................................................................................................31
Solution 27.................................................................................................................................31
Solution 28.................................................................................................................................31
Solution 29a...............................................................................................................................31
Solution 29b...............................................................................................................................31
Solution 29c...............................................................................................................................32
Solution 30a...............................................................................................................................32
Solution 30b...............................................................................................................................32
5 | P a g e

Numerical
Solution 31a...............................................................................................................................32
Solution 31b...............................................................................................................................33
Solution 32a...............................................................................................................................33
Solution 32b...............................................................................................................................33
Solution 33a...............................................................................................................................33
Solution 33b...............................................................................................................................33
Solution 34.................................................................................................................................34
Solution 35a...............................................................................................................................34
Solution 35b...............................................................................................................................34
Solution 36a...............................................................................................................................34
Solution 36c...............................................................................................................................35
Solution 36d...............................................................................................................................35
Solution 37a...............................................................................................................................35
Solution 37b...............................................................................................................................35
Solution 37c...............................................................................................................................35
Solution 38a...............................................................................................................................35
Solution 38b...............................................................................................................................35
Solution 38c...............................................................................................................................36
Solution 39.................................................................................................................................36
Solution 40a...............................................................................................................................36
Solution 40b...............................................................................................................................37
Solution 41a...............................................................................................................................37
Solution 41b...............................................................................................................................37
Solution 41c...............................................................................................................................38
Solution 41d...............................................................................................................................38
Solution 41e...............................................................................................................................38
Solution 42a...............................................................................................................................38
Solution 42b...............................................................................................................................38
Solution 42c...............................................................................................................................39
Solution 42d...............................................................................................................................39
Solution 42e...............................................................................................................................39
6 | P a g e

Numerical
Solution 43.................................................................................................................................39
Solution 44a...............................................................................................................................39
Solution 44b...............................................................................................................................40
Solution 45a...............................................................................................................................40
Solution 45b...............................................................................................................................40
Solution 45c...............................................................................................................................40
Solution 46a...............................................................................................................................40
Solution 46b...............................................................................................................................41
Solution 46c...............................................................................................................................41
Solution 47a...............................................................................................................................41
Solution 47b...............................................................................................................................42
Solution 48.................................................................................................................................42
Solution 49.................................................................................................................................42
Solution 50a...............................................................................................................................43
Solution 50b...............................................................................................................................43
Solution 51a...............................................................................................................................43
Solution 51b...............................................................................................................................43
Solution 52a...............................................................................................................................44
Solution 52b...............................................................................................................................44
Solution 52c...............................................................................................................................44
Solution 53a...............................................................................................................................44
Solution 53b...............................................................................................................................45
Worksheet Three – Kinematics.....................................................................................................45
Solution 1a.................................................................................................................................45
Solution 1b.................................................................................................................................45
Solution 2a.................................................................................................................................45
Solution 2b.................................................................................................................................45
Solution 3...................................................................................................................................46
Solution 4...................................................................................................................................46
Solution 5...................................................................................................................................46
Solution 6...................................................................................................................................47
7 | P a g e

Numerical
Solution 7a.................................................................................................................................48
Solution 7b.................................................................................................................................49
Solution 8...................................................................................................................................49
Solution 9...................................................................................................................................50
Solution 10.................................................................................................................................51
Solution 11.................................................................................................................................52
Solution 12.................................................................................................................................52
Solution 13a...............................................................................................................................53
Solution 13b...............................................................................................................................53
Solution 14.................................................................................................................................53
Solution 15.................................................................................................................................54
Solution 16.................................................................................................................................54
Solution 17.................................................................................................................................55
Solution 18a...............................................................................................................................55
Solution 18a...............................................................................................................................56
Solution 19a...............................................................................................................................56
Solution 19b...............................................................................................................................57
Solution 20.................................................................................................................................57
Solution 21a...............................................................................................................................58
Solution 21b...............................................................................................................................59
Solution 22a...............................................................................................................................59
Solution 23.................................................................................................................................60
Worksheet Four – Kinematics.......................................................................................................61
Solution 52a...............................................................................................................................61
Solution 52b...............................................................................................................................61
Solution 52c...............................................................................................................................61
Solution 53a...............................................................................................................................61
Solution 53b...............................................................................................................................62
Solution 54a...............................................................................................................................62
Solution 54b...............................................................................................................................62
Solution 54c...............................................................................................................................62
8 | P a g e

Numerical
Solution 55.................................................................................................................................63
Solution 56.................................................................................................................................63
Solution 57.................................................................................................................................64
Solution 58.................................................................................................................................65
Solution 59a...............................................................................................................................66
Solution 59b...............................................................................................................................66
Solution 60a...............................................................................................................................67
Solution 60b...............................................................................................................................68
Solution 61a...............................................................................................................................68
Solution 61b...............................................................................................................................68
Solution 61c...............................................................................................................................68
Solution 62.................................................................................................................................69
Solution 63a...............................................................................................................................70
Solution 63b...............................................................................................................................70
Solution 64a...............................................................................................................................71
Solution 64b...............................................................................................................................71
Solution 65a...............................................................................................................................71
Solution 66.................................................................................................................................72
Solution 67.................................................................................................................................72
Solution 68a...............................................................................................................................73
Solution 68b...............................................................................................................................73
Solution 68c...............................................................................................................................73
Solution 68d...............................................................................................................................73
Solution 69a...............................................................................................................................73
Solution 69b...............................................................................................................................74
Solution 69c...............................................................................................................................74
Solution 69d...............................................................................................................................74
9 | P a g e

Numerical
10 | P a g e

Numerical
Solution of Assignment_3 question
Solution 1a
As given in question,
Y = -h, a = -9.8 m/s, t =?, S = -2.0 m, u = 4.5 m/sec
We know that distance travelled is given by
S=ut+ 1
2 a t2
⟹−2=4.5 t−4.9t2
⟹ 4.9 t2−4.5 t−2=0 By solving these equations, we get,
T = 1.245 and -0.3276,
Time is non-negative, therefore, her feet in the air = 1.245 sec
Solution 1b
The highest point is calculated through v2−u2=2 aS
Putting the requied value from question (a)
s= v2−u2
2 as Putting the value in this equation,
s= 02−4.52
2(−9.8)=¿1.03316 m Ans
Solution 1c
The velocity at water level = V =u +at
⇒ v2=4.5−9.8 x 1.245 = -0.70701 m/sec
11 | P a g e

Numerical
Solution 2a
As given in question, V = 80 m/sec, g = 10 m/sec2, u = 0
We know that v2−u2=2 gh
Then, Putting the value
h= 802
2 x 10 = 6400
20 =320 m Ans
Solution 2b
Suppose the ball started from ground h = 0, Then time taken to reach the ball at the height of 320
m
S=ut+ 1
2 g t2
320=80 x t−5 t2
5 t2 −80 t+320=0
Then t = 8 sec
Time taken to reach to the ground = 8*2 = 16 sec Ans
Solution 2c
, u = 80 m/sec, g = 10 m/sec2, u = 0, t = 4 sec
The V =u +¿
Putting the value = S=80−10∗4=80−40=40 m/ sec
Solution 2d
The maximum acceleration of the ball at maximum height,
v2=u2 +2 gH
02 =802−2∗g∗320
h=6400
640 =10 m/ s2
The acceleration will be 10 m/s2 downward
12 | P a g e

Numerical
Solution 2e
Since there is no external force acting on ball after release, therefore, acceleration = 10m/sec2
upward or downward. The graph is as follows
0 2 4 6 8 10 12
-15
-10
-5
0
5
10
15
Acc vs time
Solution 2f
Since initial velocity is constant, the final velocity is changing with time, The graph of final
velocity vs time is given below
0 1 2 3 4 5 6 7 8 9
0
10
20
30
40
50
60
70
80
Velocity Time graph
Solution 2g
0 2 4 6 8 10 12
0
50
100
150
200
250
300
350
Disp Vs Time
13 | P a g e

Numerical
Solution 3a
Suppose the value of acceleration due to gravity is taken as 9.8 m/sec2
H = 30 m, T = 1 second,
We know that, h=ut + 1
2 g t2
Putting the value in it.
h=0+ 1
2∗9.8∗1=4.9m
The distance travelled in 1 second = 4.9 m
Solution 3b.
u= 0, g = 9.8 m/s2, h = 30m,
As per equation v2=u2 +2 gh
v2=0+2∗9.8∗30
v2=588 , then, v = 24.25 m/sec
Solution 3c.
First, we must calculate time taken to reach the ground
h=ut + 1
2 g t2
Putting the values
30=+1
2 ∗9.8∗t2 t = 2.47
Distance covered in 1.47 sec = 0.5*9.8*1.47 = 10.58 m
Then, distance covered = 30.0-10.58 = 19.41 m
Solution 4a.
The relative velocity of 2nd runner = 4.5-3.75 = 0.75 m/sec
Solution 4b.
Time taken by first runner = 250/3.75 = 66.66 sec
Time taken by second runner = (250+50)/4.5 = 66.66 sec
The time taken to cover their respective distance is equal to 66.66 sec, Therefore, both
runners reach at the same time
14 | P a g e

Numerical
Solution 5a.
Since both the runner, move reach the finish time at same time, so no gap between them
at finish line.
N
W E
S
As given in the figure,
Suppose the velocity of jet w.r.t air is given as Vja
v ja=250 cos ( 6o ) (−i)+ 250 sin ( 6o ) (− j)
And velocity of air w.r.t. ground = Vag
vag=30 cos ( 10o ) (−i)+30 sin ( 10o ) (− j)
Now velocity of plane w.r.t ground
v jg=v ja+ vag
Then Vjg = −250 cos ( 6o ) i−250 sin ( 6o ) j+30cos ( 10o ) i−30 sin ( 10o ) j ….(i)
Vjg = -219.081i-31.333j
Solution 5b.
The magnitude of this vector is resultant speed = 221.31 m/sec
The direction = tan−1 −31.33
−219.081 =8.14o South of west Ans
15 | P a g e
6o
10o

Numerical
Solution 6a.
Solution 6b.
As given in figure, the equilibrium of motion can be given as for horizontal direction
FCos 30−Fr=ma
⟹ FCos30o −50=30∗a
The constant velocity means, force will be applied only for friction , and acceleration is zero
FCos 30−Fr=0
FCos 30=50 Or F = 50/0.866 = 57.73 N
Worksheet One – Measurement and Mensuration
Solution 1a
100 km/hour = 1000*100/3600 = 27.77 m/sec
Solution 1b
100 km/hour = 100/1.609 = 62.1371 miles/hour
16 | P a g e

Numerical
Solution 2a
33 m/s = 33*3.6 = 118.8 km/hour
Solution 2b
The result is 118.8 km/h, therefore, it is exceeding 90km/h
Solution 3
Since we know that,
Since, distance covered in 1 sec = 1 m
Then, distance covered in 3600 sec = 3600 m
Distance covered in 1 hr = 3600m = 3.6 km
Therefore, 1.0 m/s = 3.6 km/h
Solution 4
1 yard = 3ft, and 3.281 ft = 1 m
Then 100 yard = 300ft = 300/3.281 = 91.435 m Ans
Solution 5
Length of the Soccer field = 115 m = 115*3.281 = 377.315ft = 377 ft and 0.315*12 =
3.78 inch
Length of the Soccer field = 377’ 3.78” Ans
Width of the Soccer field = 85 m = 85 x 3.281 = 278.885 ft = 278 ft and 0.885*12 =
10.61 inch
Width of the Soccer field = 278’10.61”
Solution 6
6 ft 1inch = 72+1 = 73 inch,
∴ 39.37 inch=1, then73 inch= 73
39.37 =1.8542 m
Solution 7
As given in question,
3281 ft = 1 km
17 | P a g e

Numerical
Then, 29028 ft = 29028/3281 = 8.8473 km
Solution 8
As given in question
Distance in 1 sec = 342 m/s
3600 sec = 342*3600 = 1231200 m = 1231200/1000 = 1231.2 km /hr Ans
Solution 9a
As given in question
1 year = 24*364*60*60 seconds move to 4 cm
1 sec = 4/3153600 = 126839 x 10-7 cm /sec Ans
Solution 9b
As given in question,
In 1 year moves 0.00004 km
Then 1000000 yrs. moves = 0.00004*100000 = 40 km/million-year Ans
Solution 10
As given in figure, distance between earth and sun = r = 108 = 10000000 km
Assuming as circular orbit, the distance covered by earth in 1 revolution ¿ D=2 πr
Putting the values,
D=2∗3.14∗108 =628000000 km
Distance travelled by earth in 1 year (31536000 sec) = 628000000 km
Distance travelled by earth in 1 second = 628000000
31536000 =19.91375 km/sec Ans
Solution 11
The uncertainty for a quantity is given as
δa= %δa
100 x quantity= 3
100 x 65=1.95 kg
18 | P a g e

Numerical
The uncertainty of mas in 65 kg is 1.95 kg
Solution 12
The % uncertainty in measuring δl
L x 100= 0.005
20 x 100 (0.5 cm = 0.005 m)
δl
L x 100=0.025 % Ans
Solution 13
The uncertainty for a quantity is given as
δa= %δa
100 x quantity= 5
100 x 90=4.5 km
The uncertainty of mas in 90 km is 4.5 km, which means the actual speed is in the range
of 90-4.5 = 85.5 km/h or 90+4.5 = 94.5 km/h
Or, range is 85.5*0.6214 = 53.13 miles/hour to 58.72 miles/hour
Solution 14
The % uncertainty in measuring 5
130 x 100= 5
130 x 100
δl
L x 100=± 3.84 % Ans
Solution 15
As given in question, 72 beats /min = 72/60 = 1.2 beats /sec (Since 1 year = 31536000 sec)
In 2.0 years, the no of beats = 31536000*1.2*2 =75686400 = 7.5 x 107 beats
In 2.00 years, the no of beats = 75.68x 106 beats
In 2.000 years, the no of beats = 7.568 x 107 beats
19 | P a g e

Numerical
Solution 16
The Soda left after 308 ml removal = 375-308 = 67 mL
Solution 17a
( 106.7 ) ( 98.2 )
( 46.21 ) ( 1.01 ) In the given equation the lowest number of significant figures is 3.
Therefore, answer should be 3 significant figures
Solution 17b
(18.7)2 Only one number which has significant figure 3. Therefore, solution will be 3
significant figures.
Solution 17c
(1.60 x 10-19) (3712) = The lowest significant figure is 3, Therefore solution should be 3
significant figures.
Solution 18a
The number 99 consists of 2 significant figures, and 100 has 3 significant figures
Solution 18b
The uncertainty in 99 = 1
99 x 100=1.0 %
The uncertainty in 100 = 1
100 x 100=1.00 %
Solution 18c
As per result obtained from part a and b, The % uncertainty show deviation from the
results, therefore, % uncertainty is more meaningful.
Solution 19a
The % uncertainty in speedometer = δl
L x 100= 2
90 x 100 = 2.22%
Solution 19b
The range of speed at 60 km/h = δ a= %δa
100 x quantity= 2.22
100 x 60=1.33 km/h
The range of the speed in between = 55.667 to 61.333 km/h Ans
20 | P a g e

Numerical
Solution 20a
The % uncertainty in blood pressure= ± 2
120 x 100= 2
120 x 100 = 1.667%
Solution 20b
The uncertainty in blood pressure at the 80 mmHg = 80*1.667/100 = ±1.33 mm Hg
Solution 21
The heart rate for one minute = 40∗60
30 =80 beats/min
The % uncertainty in heart rate= 1
40 x 100+ 0.5
30 x 100 = 2.5% + 1.7% = 4.2%
Therefore, uncertainty in heart rate
δa= %δa
100 x quantity= 4.5 %
100 x 80=3 beats/min
Therefore, heart rate is 80 beats/min with uncertainty 3 beats/min
Solution 22
The area of circle is given by = π ¿ Then putting the values
3.14 x 3.1022
4 =7.5536 c m2 Ans
Solution 23
The time taken to cover the marathon at the speed of 9.5 mi/h 26.22
9.5 =2.76 h Ans
Solution 24
The distance in meter = 42188 m
Solution 24a
The uncertainty in distance = %δ= δ
distance = 25
42188∗100=0.05926 %
Solution 24b
Time in seconds = 2*3600+30*60+12= 9012 sec Ans
21 | P a g e

Numerical
The uncertainty in time = %δ= δ
distance = 1
9012∗100=0.0111 % Ans
Solution 24c
The average speed in meter / sec = 42188/9012= 4.681 m/sec Ans
Solution 24d
In this condition we must take two averages as per uncertainty, one is max and other is
min average.
Max average 42188+25
9012−1 =4.684 m/sec
Min Average = 42188−25
9012+1 =4.678 m/sec Ans
Solution 25
The volume of the box can be given as V =xyz (Putting the value)
Volume of box = 1.8 x 2.05x3.1 = 11.439 cm3
The max volume with uncertainty = (1.8+0.01) (2.05+0.02) (3.1+0.1)
Vmax = 11.99 cm3
The min volume with uncertainty = (1.8-0.01) (2.05-0.02) (3.1-0.1)= Vmin = 10.90 cm3
The volumetric with uncertainty = 11.99-10.90 = 1.09 cm3 Ans
Solution 26
Since 1 pound = 0.4539 kg
Then, 0.0001kg = 0.0001/0.4539 = 0.00022 lb
The % uncertainty in pound = 0.00022/1 % = 0.022 %
The uncertainty in 1kg = 1000 gm = 1000*0.022 = 22 gm Ans
Solution 27
The area of room A = lxb = 3.955*3.050 = 12.063 cm2
The Max area of room with uncertainty = (3.955+0.005) (3.05+0.005) = 12.098
The min area of room with uncertainty = (3.955-0.005) (3.05-0.005) = 12.028
22 | P a g e

Numerical
The amount of uncertainty in area = 12.098-12.028 = 0.070 cm2 Ans
Solution 28a
The volume of piston = π ( d
2 )2
l=3.14 x 7.52 x 3.25
4 =143.508 c m3
The gas decreases in piston in volume = 143.508 cm3
Solution 28b
The max volume with uncertainty = 3.14 x ( 7.5+0.002 )2 ( 3.25+0.001 )
4 = 143.63 cm3
The max volume with uncertainty = 3.14 x ( 7.5−0.002 )2 ( 3.25−0.001 )
4 = 143.39 cm3
Uncertainty in volume = 143.63-143.39 = 0.2414 cm3
Worksheet Two – Kinematics
Solution 1a
The distance covered by path A = Δ x=|xf −xi|=7−0=7 m Ans
Solution 1b
The magnitude of the distance covered = | Δ x|=|xf −xi|=|7 m|=7 m Ans
Solution 1c
The displacement from start to finish = Δ x=x f −xi=7−0=+7 m Ans
Solution 2a
The distance covered by path B = Δ x=|xf −xi|=|7−12|=5 m Ans
Solution 2b
The magnitude of the distance covered = |Δ x|=|xf −xi|=|7−12|=5 m Ans
Solution 2c
The displacement from start to finish = Δ x=x f −xi=−7 +12=−5 m Ans
23 | P a g e

Numerical
Solution 3a
The distance covered by path C
= Δ x=|xf 1
−xi2
|+|xf 3
−xi4
|+|xf 5
−xi6
|
¿|10−2|+|8−10|+|10−8|=8+2+2=12 m Ans
Solution 3b
The magnitude is given by = Δ x=|11−2|=9 m Ans
Solution 3c
The displacement from start to finish = Δ x=x f −xi=11−2=9 m Ans
Solution 4a
The distance covered by path D
= Δ x=|xf 1
−xi2
|+|xf 3
−xi4
|=|9−3 |+|5−3|=6+2=8 m Ans
Solution 4b
The magnitude of the distance covered = | Δ x|=|−6+ 2|=4 m Ans
Solution 4c
The displacement from start to finish = Δ x=x f −xi=−6+2=−4 m Ans
Solution 5a
As given in figure, distance between earth and sun = r = 108 = 10000000 km
Assuming as circular orbit, the distance covered by earth in 1 revolution ¿ D=2 πr
Putting the values,
D=2∗3.14∗108 =628000000 km
Distance travelled by earth in 1 year (31536000 sec) = 628000000 km
Distance travelled by earth in 1 second = 628000000
31536000 =19.91375 km/sec
Solution 5b
Since the average velocity in a year = displacement /time,
24 | P a g e

Numerical
But in case of revolution displacement is zero, therefore, average velocity = 0
Solution 6
Since the average velocity in a year = displacement /time,
But in case of revolution displacement is zero, therefore, average velocity = 0
Solution 6
As given in question, 100 revolution /min = 100/60 = 5/3 rev /sec
The circumference of the radius = 2 πr = 2 *3.14*5 = 31.4 m
Distance covered in 1 sec = speed = 31.4*5/3 = 52.33 m/sec Ans
Solution 7
500 km = 500*1000*100 cm
Speed of drifting = 3 cm/year
Time taken to cover 5.0*107 cm = 5/3 * 106 = 1.66 x 107 year Ans
Solution 8
590 km = 590*1000*100 cm
Speed of drifting = 6 cm/year
Time taken to cover 5.90*107 cm = 5/3 * 106 = 9.833 x 106 year Ans
Solution 9
Total time in hour = 13h + 4/60 h+ 58/3600 hour = 13.08278 h
Average speed of train = distance / time = 1633.8/13.08278 = 124.88 km/h
Speed in m/sec = 124880/3600 = 34.68 m/sec
Solution 10
We know that, 3.84 x 106 m = 384x106 cm
Time taken to cover this distance = 384∗106
4 =¿96 x 106 year Ans
25 | P a g e

Numerical
Solution 11a
The average speed of the car = 12*60/18 = 40 km/hour
Solution 11b
The average velocity = displacement / time = 10.3 *60/18 = 34.33 at the direction 25o
south of east.
Solution 11c
Total time by trip = 7.5 hr
Overall speed of return journey = 24/7.5 = 3.2 km /hr.
Solution 11d
Since the overall displacement is zero, therefore, average velocity will also be zero
Solution 12
Time taken to travel the nerve cell = 1.1/18 = 0.0611 sec Ans
Solution 13
Distance covered by echo time = distance between earth and moon = speed x time
Distance between earth and moon = 2.56 x 3 x108 = 7.68 x 108 meter Ans
Solution 14a
The velocity for straight down 15 m = 15/2.5 = 6
The velocity of backward 3 m =3/1.75 = 1.71 m
The velocity of straight forward = 21/5.2 = 4.0384 m/sec
Solution 14b
Total displacement = 15 -3+21 = 33 m
Total time = 2.5 + 1.75 + 5.2 = 9.45 sec
Then average velocity = 33/9.45 = 3.49 m /sec in forward direction
Solution 15a
Circumference of the hydrogen atom ¿ πd=¿ = 3.14 x 1.06 x 10-10 m,
Speed of the electron = 2.2 x 10-10 m/s
26 | P a g e

Numerical
Time in one revolution = πd /v = 1.5129 x 10-16 sec
Revolution in one sec = 1/(1.5129 x10-15) = 6.61 x 1015
Solution 15b
Since displacement is zero after each revolution, then velocity will also be zero.
Solution 16
We know that acceleration = v-u/t = 30-0/7 = 4.286 m/sec2 Ans
Solution 17a
For first case, v = 282 m/s, u = 0, t = 5 s
Acceleration = 282-0/ 5 = 56.4 m/sec2 = 56.4/9.8 = 5.755 g
Solution 17b
For second case, u = 282 m/s, v = 0, t = 1.4
Then deceleration = 0-282/1.4 = -201.42 m/sec2 = -201.42/9.8 = 20.55g Ans
Solution 18a
As given in question, v = 2.00 m/s, u = 0, a = 1.4 m/s2
V =u+at = 0+1.4t, or t = 2.00/1.4 = 1.4285 seconds
Solution 18b
U = 2.00, v = 0, t = 0.8 sec
Then a = v-u/t = 0-2/0.8 = -2.5 m/s2 Ans
Solution 19
As given in question, 6.5 km/s = 6.5*1000 = 6500 m/sec t = 60 s, u = 0
Then acceleration = v-u/t = 6500-0/60 = 108.33 m/s2
The average acceleration of ballistic missile = 108.33/9.8 = 11.054 g Ans
Solution 20
As given in question,
A = 4.50 m/s2, t = 2.4 s, u = 0
27 | P a g e

Numerical
Then v = u+at , v = 0+4.5*2.4 = 10.8 m/s2 Ans
The graph of distance vs time is given below
0 0.5 1 1.5 2 2.5 3
0
5
10
15
20
Distance Vs Time
time(s)
Distance (m)
Ans
Solution 21
As given in question,
v = 0, a = -2.1 x 104 m/s2, u = ?, t = 1.85 x 10-3 seconds
We know that, v = u+at, Putting the values,
0 = u – 2.1 x 104 x 1.85 x 10-3 or, u = 2.1 x 104 x 1.85 x 10-3 = 38.85 m/s
Velocity when touches the mitt = 38.85 m/s
Solution 22
As given in question,
U = 0, t = 8.1 x 10-4, a = 6.2 x 105 m/s2, v = muzzle velocity = ?
We know that v = u+at, then v = 0+ 8.1*10-4 x 6.2x105 = 502.2 m/s
Therefore muzzle velocity = 502.2 m/s
Solution 23a
As given in question,
U = 0, t =?, a = 1.35 m/s2, v = 80*1000 m /h = 22.22 m/s
28 | P a g e

Numerical
We know that v = u+at, then 22.22 = 0 + 1.35*t
Time taken to accelerate = t = 22.22/1.35 = 16.46 sec
Solution 23b
As given in question,
U = 22.22, t =?, a = -1.65 m/s2, v = 0
V = u+at, t = v-u/a = -22.22/-1.65 = 13.46 seconds
Solution 23b
During emergency
U = 22.22, t = 8.3, v = 0, a =?
We know that, a= v−u
t putting the values,
a=−22.22
8.3 =−2.67 m/s2
In case of emergency train decelerates at -2.67 m/2
Solution 24a
In this problem,
Solution 24b
The known variable are listed below
U = 12 m/s, a = 2.4 m/s2, t = 12 seconds, and unknown is final velocity v
Solution 24c
The unknown variable is final velocity, which can be calculated as follows
29 | P a g e

Numerical
v=u+at , Putting the values
V = 0+2.4 x 12 = 28.8 m/sec.
After knowing the unknown variable we can calculate the distance travelled in 12
seconds, which is as follows
v2=u2 +2 aS Where S is the distance travelled in meter. Putting the value in
above equation,
28.82 = 02+2 x 2.4 x S
Or, S = 28.82/4.8 = 172.8 m. therefore, distance travelled = 172.8 m
Solution 25a
U = 9.00 m/s, a = 2.00 m/s2, t = 5.00 sec
Distance travelled in 5 sec = ut +0.5at2 = 9*5-0.5*2*52 = 45-25 = 20 m
Therefore, distance travelled by runner = 20 m
Solution 25b
Final velocity after 5 sec, v = u+at, v = 9-2*5 = 9-10 = -1 m/sec
The calculation shows that, final velocity is negative, we must calculate the time at final velocity
0
T = v-u/a = 0-9/2 = 4.5 sec,
Solution 25c
The runner stops at 4.5 sec, therefore final velocity of the runner must be zero after 5 sec
Ans
Solution 26a
The sketch is given below
30 | P a g e

Numerical
Solution 26b
The known variable in this problem is listed below
U = 0, v = 30 cm/s, S = 1.8 cm
Solution 26c
As given above known variable,
v2=u2 +2 aS Putting the value in this equation
302 = 0+2 x a x 1.8
Then, a = 900/3.6 = 250 cm/sec2,
Then time taken to accelerated = t = 30/250 = 0.12 seconds
Solution 26d
Since our heart rate is 72 beats in 60 seconds, i.e. one beat is in 0.83 seconds, The time of
blood acceleration is only a fraction of time, therefore, answer is reasonable.
Solution 27
U = 8 m/s, v = 40 m/s, t = 3.33 x 10-2 s, a = ?, S = ?
We know that, a = (v-u)/t = (40-8 )/3.33x10-2 = 960.961,
The distance covered during acceleration
v2=u2 +2 aS Putting the value in this equation
= 402 = 82 + 2 x 960.961xS
Then S = 1536/1921.922 = 0.7992 m Ans.
Solution 28
U = 0, v = 26.8 m/s, t = 3.9 s, a =?
The average acceleration of bike = 26.8-0/3.9 = 6.87 m/s2
Distance travelled during acceleration
S = 0.5 x 6.87 x 3.92 = 52.25 m Ans
31 | P a g e

Numerical
Solution 29a
As given in question,
U = 4 m/s, a = 0.05m/sec2, v = ?, t= 8*60 = 480 sec
We know that, v=u+at, Putting the value in eq.
V= 4+480*0.05 = 24+4 = 28 m/sec
Final speed of train is 28 m/s
Solution 29b
U = 28 m/s, v = 0, a = .55m/s, t = ?
We know that, (v−u)/t =( 28-0)/0.55 = 50.9090 seconds
Time taken to stop the train 50.9090 seconds
Solution 29c
Case 1 Acceleration,
Now, u = 4 m/s, t = 480 s, a = 0.05
We know that S=ut+ 1
2 a t2
S = 4 * 480 + 0.5 x 0.05x4802 = 7680 m
Case 2 deceleration,
Now, u = 28 m/s, t = 50.91 s, a = 0.55
We know that S=ut+ 1
2 a t2
S = 50.91 * 28 - 0.5 x 0.55x50.912 = 712.7273 m
Solution 30a
U = 0, v = 65.0 m/sec, S = 0.250 m, a =?, t = ?
v2=u2 +2 aS Putting the value in this equation
= 652 = 02 + 2 x 0.25xa
a = 4225/0.5 = 8450 m/s2
32 | P a g e

Numerical
U = 0, v = 65.0 m/sec, S = 0.250 m, a = 8450 m/s2, t =?
T = (v-u)/a = 65/8450 = 0.007692 s Ans.
Solution 30b
U = 0, v = 65.0 m/sec, S = 0.250 m, a =?, t = ?
v2=u2 +2 aS Putting the value in this equation
= 652 = 02 + 2 x 0.25xa
a = 4225/0.5 = 8450 m/s2
Solution 31a
U = 0, v = 6 m/s, a = 0.35 m/s2
Distance travelled in accelerating v2=u2 +2 aS
62 = 02 + 2x0.35xS
Then S = 36/0.7 = 41.42 m
Solution 31b
V = 6 m/s, u = 0, a = 0.35 m/s2, t =0
T = (v-u)/a = 6/0.35 = 17.14 seconds
Therefore, time taken to become airborne is 17.14 seconds and distance covered is 41.42 m
Solution 32a
As given in question,
U = 0.60 m/s, v = 0, S= 2.00 mm = 2 x 10-3 m
We know that, a= v2−u2
2 S = −0.62
2 x 2 x 10−3 = -90 m/s2
The deceleration in g format = 90/9.8 = 9.184 g
Solution 32b
Now time can be calculated as
33 | P a g e

Numerical
t= 2 s
u+v Putting the values
¿ 2 x 2 x 10−3
0.6 =¿ 0.00667 seconds Ans
Solution 33a
U = 7.5 m/s, v = 0, S= 0.35 m,
We know that, a= 0−u2
2 S = −7.52
2 x 0.35 = -80.36 m/s2 Ans
Solution 33b
Now time can be calculated as
t= 2 s
u+v Putting the values
¿ 2 x 035
7.5 =¿ 0.0933 seconds Ans
The deceleration of player -80.36 and time to collide = 0.0933 s
Solution 34
As given in question,
U = 54m/s, g = -9.8 m/sec2, v = 0, t=? Sec, S = 3 m
We know that, a= v2−u2
2 S =−542
2∗3 = -486 m/s2
Solution 35a
As given
V = ?, u = 0, g = 9.8 m/s2, h = 3 m
v2=u2 +2 aS Putting the value in this equation
V2 = 2 x 9.8 x 3 , then v = 7.668 m/s
Solution 35b
Now, u = 7.668 m/sec, v = 0, S = 2/100 = 0.02 m
34 | P a g e

Numerical
We know that, a= v2−u2
2 S =−7.6682
2 x 0.02 = -1469.96 m/s2
Therefore, squirrel decelerates at 1469.96 m/s2
Solution 36a
As given, u = 22 m/s, a =- 0.15 m/s2, S =210 m,
The time taken to travel the nose of the train will be distance covered by engine on the
platform
S = ut +0.5at2 Putting the value in equation
210 = 22t-0.5x0.15t2,
Then 0.075t2 – 22t+210 then t = 9.8781, 283.46
Therefore, t = 9.8781 and 283.46 Ans.
To decide the time of crossing the train, we must find the time to become stops the train
Using V = u + at, Then, 0 = 22-0.15*t, or t = 22/0.15 = 146.667.
Since the train stops at 146.667 seconds, therefore, train must cross the platform at time 9.87 s
Solution 36c
The distance travelled by train must include platform + length of train = 130+210 = 340m
We will use the following station
S=ut+ 1
2 a t2
Putting the values, 340 = 22t- 0.5*0.15*t2
Or, 0.075t2 -22t +340 = 0 , T = 16.37 seconds Ans
Solution 36d
U = 22 m/sec, S = 340 m, a = -0.15 m/s2
Using the formula, v2=u2 +2 aS
v2=222−2∗0.15∗340 ⟹ v=19.54 m/s Ans
35 | P a g e

Numerical
Solution 37a
The average acceleration of the dragster ¿ v−u
t =145−0
4.45 =32.58 m/s2
Ans
Solution 37b
As information obtained from (a)
U = 0 m/se, S = 402 m, v =?, a = 32.58 m/s2
v2=02 +2 a S = + 2 *32.58*402
Then v = 161.8466 m/sec Ans
Solution 37c
For any vehicle, the change of gear reduces the acceleration of the vehicle, same has
happened to dragster also. The final velocity would be less that what calculated above.
Solution 38a
U = 11.5 m/s, a = 0.5 m/s2, t = 7 sec
V =11.5+ 0.5∗7=15 m/sec
Solution 38b
Time taken to cover 300 m at constant velocity 11.5 m/s = 300/11.5 = 26.09 s
Distance covered at 0.5 m/s2 acceleration = S = 11.5 x 7 +0.5x0.5*7 = 92.75
The distance covered by racer in 7 seconds is 92.75
The rest distance covered by constant velocity of 15 m/sec = 300-92.75 = 13.82
Total time taken by racer = 13.82+7 = 20.82
Saved time = 26.09-20.82 = 5.27 Seconds
Solution 38c
Th time taken by 2nd racer = 295/11.8 = 25.0 s
Time taken by first racer to cover 300 m = 20.82 sec
The time difference of winner racer = 25-20.82 = 4.18 seconds Ans
The difference in distance = 4.18*15 = 62.7 m
36 | P a g e

Numerical
Solution 39
The acceleration can be calculated as,
a=(v−u)/t=(60−0)/ 4=15 m/ s2
The time required to attain the speed = 183.58 mi/h
t= 183−0
60 =12.2 s
The distance travelled during acceleration
s= vm+ u
2 t= 183.58
2 x 12.2
3600 =0.31 mi
The distance remained to cover = 5.0-0.31 = 4.7 mi
And Time for rest of the distance = 4.7
183.58 x 3600=92.17 s
Then total time required to cover the distance = 92.17 + 12.2 = 104.37 seconds
Solution 40a
A given in question u = 0, t = 9.69 sec, time for acceleration = 3 s, S = 100 m
The distance covered in constant speed = d= v+u
t t= 3 xv
2 =1.5 v
If we substitute the value of d for velocity part
100−d=6.69 v Putting the value d =1.5v
100-1.5v= 6.69v
Or, v = 100/8.19 = 12.21 m/s
Now we got final velocity v =12.21 m/s
Then acceleration
a= v−u
t =12.21−0
3.0 =4.07 m/ s2 Ans
37 | P a g e

Numerical
Solution 40b
In this case the decelerating part is = 200-d
The covered distance = d = v +u
2 t= v
2 x 3=1.5 v
Equating the problem as previous
200 – d = v(16.30)
Or, v = 200/17.88 = 11.2 m/se
Therefore, maximum velocity of Usain Bolt = 11.2 m/se
Solution 41a
As given in question,
U = 15.0 m/s, g = -9.8 (Upward), t = 0.5
The attained height H = 15*0.5 – 0.5*9.8*.52 = 7.5 – 1.225 = 6.275 m
Final velocity v = 15-9.8*0.5 = 15-4.9 = 10.1 m/s
Solution 41b
As given in question,
U = 15.0 m/s, g = -9.8 (Upward) t = 1.00
The attained height H = 15*1 – 0.5*9.8*12 = 15 – 4.9 = 10.1 m
Final velocity v = 15-9.8*0.5 = 15-4.9 = 5.2 m/s
Solution 41c
As given in question,
U = 15.0 m/s, g = -9.8 (Upward) t = 1.50
The attained height H = 15*1.5 – 0.5*9.8*1.52 = 22.5 – 11.025 = 11.475 m
Final velocity v = 15—14.7= 0.3 m/s
Solution 41d
As given in question,
38 | P a g e

Numerical
U = 15.0 m/s, g = -9.8 (Upward) t = 2.0
The attained height H = 15*2 – 0.5*9.8*22 = 30 – 19.6 = 10.4 m (falling)
Final velocity v = 15—19.6= -4.6 m/s (downward)
Solution 41e
As given in question,
U = 15.0 m/s, g = -9.8 (Upward) t = 2.5
The attained height H = 15*2.5 – 0.5*9.8*2.52 = 37.5 – 30.625= 6.875 m (falling)
Final velocity v = 15—24.5= -9.5 m/s (downward)
Solution 42a
As given in question,
U = 14.0 m/s, g = 9.8 (Downward), t = 0.5, H = 70 m
The attained height H = 14*0.5 + 0.5*9.8*.52 = 7 – 1.225 = 8.225 m
Position of stone = 70-8.225 = 61.775 m above ground
Final velocity v = 14+9.8*0.5 = 14+4.9 = 18.9 m/s
Solution 42b
As given in question,
U = 14.0 m/s, g = 9.8 (Downward), t = 1.0, H = 70 m
The attained height H = 14*1 + 0.5*9.8*12 = 14 + 4.9 = 18.9 m
Position of stone = 70-18.9 = 51.1 m above ground
Final velocity v = 14+9.8*0.5 = 14+9.8 = 23.8 m/s
Solution 42c
As given in question,
U = 14.0 m/s, g = 9.8 (Downward), t = 1.5, H = 70 m
The attained height H = 14*1.5 + 0.5*9.8*1.52 = 32.025 + 37.975 = 37.975 m
Position of stone = 70-32.025 = 37.975 m above ground
39 | P a g e

Numerical
Final velocity v = 14+9.8*1.5 = 14+14.7 = 28.7 m/s
Solution 42d
As given in question,
U = 14.0 m/s, g = 9.8 (Downward), t = 2.0, H = 70 m
The attained height H = 14*2.0 + 0.5*9.8*2.02 = 28 + 19.6 = 47.6 m
Position of stone = 70-47.6 =22.4 m above ground
Final velocity v = 14+9.8*2.0 = 14 19.6 = 33.6 m/s
Solution 42e
As given in question,
U = 14.0 m/s, g = 9.8 (Downward), t = 2.5, H = 70 m
The attained height H = 14*2.5 + 0.5*9.8*2.52 = 35 + 30.625 = 66.625 m
Position of stone = 70-4.375 =4.375 m above ground
Final velocity v = 14+9.8*2.5 = 14 +24.5 = 38.5 m/s
Solution 43
U = 0, g = -9.8 m/s2, H = 1.25 m
We know that, v2=u2 +2 gH Putting the value in equation
v=− √ 2∗9.8∗1.25=4.95 m/sec(Upward ) Ans
Solution 44a
The known variable is as follows
U = 1.4 m/s, t = 1.8 s, g = 9.8 m/s2
Solution 44b
H = ut+0.5t2 = 1.4x1.8 + 0.5*1.8*1.8*9.8 = 2.52 + 15.876 = 18.396 m Ans
Solution 45a
The known variable in this problem is as follows
U = 13.0 m/s, g = 9.8 m.s2, v = 0, H = ?
40 | P a g e

Numerical
The height attained by dolphin, is at hat point where final velocity gong upward will be
zero.
Solution 45b
From the equation
v2=u2−2 gH Putting the value of known variable
0 = 132 -2 *9.8*H, or H = 169/19.6 = 8.62245 m
Solution 45c
The time taken to reach the highest point t = v−u
g = −13
−9.8 =1.327
The time taken to going upward is 1.327 s, similarly time taken to going downward = 1.327
Total time in air = 1.327+1.327 = 2.653 s
Solution 46a
As given in question, u = 4 m/s, g = -9.8 m/s2(upward), H = 1.8 m, v = 0
Since The swimmer jumps from board, therefore, distance covered will addition of board height
v2=u2 +2 gH 0 = 42 – 2*9.8 *H
Or H = 16/19.6 = 0.816 m,
The total height = 1.8 m+0.816 = 2.616 m
The time taken in going up = t = 4-0/9.8 = 0.408 s
Now time taken to come to the water
S = ut+1/2gt2 Putting the value,
2.616 = 4.9t2 or, t=√ 4.9/2.616= 1.37 sec
Total time in the air = 0.408+1.37 = 1.77661 s
Solution 46b
As given in question, u = 4 m/s, g = -9.8 m/s2(upward), H = 1.8 m, v = 0
Since The swimmer jumps from board, therefore, distance covered will addition of board
height
41 | P a g e

Numerical
v2=u2 +2 gH 0 = 42 – 2*9.8 *H
Or H = 16/19.6 = 0.816 m,
The total height = 1.8 m+0.816 = 2.616 m
Solution 46c
The final velocity can be calculated as
v2=u2 +2 g H
V2 = 2 * 9.8*2.616 = 51.2736, then v = 7.1606 m/sec Ans
Solution 47a
As given, t = 2.35 s, u = 8.00 m/s, g =-9.8 m/s2 (Upward)
H = ut+0.5gt2 = 8*2-0.5*9.8*2.352 =
H = ut+0.5gt2 = 8*2+0.5*9.8*2.352 = 18.8 + 27.0602 = 8.26 m
The height of the Thrown stone from cliff = 8.26m
Time taken to reach the top point
T = -8/-9.8 = 0.816 s
The time taken by stone to reach the ground = 2.35 – 0.816 = 1.533673
Now I must calculate distance covered in 1.534 sec
H = ut +0.5gt2 = 0+0.5*9.8*1.5342 = 11.52556
Then height of the cliff = 3.265 m
Solution 47b
Time taken to reach the stone when thrown downward,
U = 8.0 m/s, H = 3.265, g = 9.8
We know that, H = ut +0.5at2, Putting the value
3.265 m = 8t+4.9t2
42 | P a g e

Numerical
4.9t2 + 8t-3.265 = 0
Calculated root is t=−400 ± √319985
490
T = 0.33811 and – 1.97076
Since negative amount is neglected for time, t = 0.33811 seconds Ans
Solution 48
As given in question,
U = 11 m/sec, g = 9.8 m/sec2, height = 2.2m, another height = 1.8 m
The equation for distance covered = H = H’ + ut +0.5at2
Putting the value
2.2 = 1.8 +11 -0.5 *9.8*t2
Solving the equation through quadratic equation
t=− ( 11 ) ± √ 112− (−9.8 ) ( 2 ) ( 2.2−1.8 )
−9.8 =0.02∨2.26 s
Negative value is omitted, t = 2.26 seconds,
Then shot putter must get away = 2.26 seconds
Solution 49
As given, u = 15 m./sec, H = 7m, g = 9.8,
The formula for distance H−H' = yt+ 1
2 g t2
With the help of above equation, the equation of time
t=
−u ± √ u2−4 (1
2 a)(h−h')
2 ( 1
2 ) a
Putting the required value from question,
t=−15 ± √ 152−2(9.8)(7−0)
−9.8
43 | P a g e

Numerical
After solving the above quadratic equation, we get t =0.57 and 2.49 s
Tome taken to pass the branch while upward = 0.57 s
Tome taken to pass the branch while upward = 2.49 s
The required time interval = 2.49 -0.57 = 1.92 s Ans
Solution 50a
As given, H = 2.5 m, v =0, u =?, g = 9.8 (upward)
v2=u2 +2 g H Putting the value
02 = u2 – 2*9.8*2.5, u2 = 49, then u = 7 m/s Ans
Solution 50b
Time taken to reach the air = t = 2H/(u+v) = 5/7 = 0.714 seconds Ans
Solution 51a
The reference plane distance = 0, Suppose the rock reaches at place H after 1.5 s
Then
H-Ho = ut+1/2gt2, y = 0*1.5 + ½(-9.8)(1.5)2 = -11.02 m
The rock will be visible for hiker H = 105 – 11.02 = 93.98 m
Solution 51b
The time taken to reach the ground by rock can e given as
t= √ 2 ( H −Ho )
a Putting the values
t= √ 2 ( 0−105 )
−9.8 =4.6 s
Time for hiker to take action = 4.6-1.5 = 3.1 seconds Ans
Solution 52a
As given in question
U = 0, t = 1 sec, g = 9.8 m/sec2 (downward)
We know that,
44 | P a g e

Numerical
, H = ut +0.5at2, Putting the value
H = 0-4.912 = -4.9 m
Distance traveled in 1st sec = 4.9 m
Solution 52b
v2=u2 +2 as
The final velocity of the object is
v=± √u2+2 as
v=± √ 2∗(−9.8)(−75)=± 38 m/ s
The velocity at which the ball reach at ground = 38 m/sec
Solution 52c
We know that,
H−H o=ut+ 1
2 a t2
y= 1
2 at2 or t= √ 2 y
a = √ 2∗75
9.8 =3.9 seconds Ans
Then distance travelled by body in first 2.9 second
H = 0 + 0.5*9.8*2.92 = 41 m
Hence distance travelled in last second = 75-41 = 34 m Ans
Solution 53a
As given, u = 0, H = 250 m, g = 9.8 m/sec2(Upward)
v=± √ 2∗9.8∗250=± 70 m/s
The final velocity while hitting the ground = 70 m
Solution 53b
Sound speed = 335 m/s, H = 250
Time taken to reach to sound =250/335 = 0.75 seconds
The reaction time by tourist = T = 0.75 + 0.30 = 1.05 seconds
45 | P a g e

Numerical
Time taken by rock to reach = t = (v-u)/a = 70/9.8 = 7.143 seconds
Then time for action taken by tourist = 7.143 – 1.05 = 6.10 seconds Ans
Worksheet Three – Kinematics
Solution 1a
As given in figure,
The distance covered by A is three blocks to the north and one block east,
Then, distance travelled by A = (120 +120+120) +(120) = 480 m Ans
Solution 1b
The displacement is measure as perpendicular = 120 + 120 +120 = 360 m
Now we must calculate hypotenuse¿ √ 3602 +1202= √129600+14400= √144000=379 m
The direction = tanθ=3 /1 , thenθ=71.56o Ans
Solution 2a
As given in figure,
Distance travelled in east direction, d1 = 4 *120 = 480 m
Distance travelled in north direction d2 = 3 *120 = 360 m
Distance travelled in west direction d3 = 3 *120 = 360 m
Total distance travelled = 480 + 360 + 360 = 1200m = 1.2 km
Solution 2b
Net movement in east = 120 m
Net movement in north = 360 m
The displacement can be given by h =√ 1202 +3602=379.5 m
The direction of the movement = tanθ= 360
120 , thenθ=71.56o Ans
46 | P a g e

Numerical
Solution 3
The displacement component in north direction can be given as
Sn= ( S ) Sinθ Putting the value
Sn = 5 x Sin40o = 5 *0.642 = 3.21 km
Solution 4
Suppose, ⃗ x=18 m ,⃗ y=25m ,
Resultant vector at 90o
|R|= √ 182 +252 +2 ABCos 90o= √182 +252=30.8 m
The angle between these two vectorstanθ=25 /18 , thenθ=54.25o
Actual angle = 90o 54.250 = 35.75o
Therefore, the magnitude and direction = 30.8 m and 35.75o west of north
Solution 5
The figure for the arrangement is as follows.
Suppose x component vector = As =A*Cosθ
Putting the value for given θ
As = 12.0 x Cos (90o +20o)
47 | P a g e

Numerical
As = 120 x Sin(20o) = 12.0 * (-0.342) = -4.104 m
For the same way, component along y axis
Ay = 12.0 x Cos20o = 12.0 x 0.939 = 11.27 m
Similarly, for B vector component along the x and y axis
Bx = 20 *Cos(220o) = 20 * (-0.667) = -15.32 m
And By = 20 *Sin(220o) = 20 * (-0.6427) = -12.85 m
Therefore, resultant force in x and y axis
Rx = -4.104 – 15.32 = -19.42 m
Ry = 11.27-12.85 = -1.58 m
The resultant magnitude
R = √ ( 19.42 ) 2+ ( 1.58 ) 2= √ 379.63=19.49 m Ans
And direction tanθ= 1.58
19.42 , thenθ=4.65o Ans
Solution 6
The diagram for the problem can be stated as
We must calculate the x component
Rx=−20 xCos ( 40o )−12.0 sin ( 20o )=−15.32−4.1=−19.42 m
48 | P a g e

Numerical
Ry =−20 xSin ( 40o ) +12.0 xSin ( 20o ) =−12.86+11.28=−1.58 m
The resultant R can be calculated as
R= √ ( −19.42 ) 2 + ( −1.58 ) 2=19.48 m
θ=tan−1
( −1.58
−19.42 )=4.65o Ans
Solution 7a
The component in Ax and Ay can be given as
Suppose x component vector = As =A*Cosθ
Putting the value for given θ
As = 12.0 x Cos (90o +20o)
As = 120 x Sin(20o) = 12.0 * (-0.342) = -4.104 m
For the same way, component along y axis
Ay = 12.0 x Cos20o = 12.0 x 0.939 = 11.27 m
Similarly, for B vector component along the x and y axis
Bx = 20 *Cos(40o) = 20 * (-0.667) = 15.32 m
And By = 20 *Sin(40o) = 20 * (-0.6427) = 12.85 m
Therefore, resultant force in x and y axis
Rx = -4.104 +15.32 = 11.21 m
Ry = 11.27+12.85 = 24.12 m
The resultant vector
The resultant magnitude
R = √ ( 11.21 )2+ ( 24.12 )2 = √707.43=26.6 m Ans
And direction tanθ= 24.12
11.21 , thenθ=65.1o Ans
49 | P a g e

Numerical
Solution 7b
The component in Ax and Ay can be given as
Suppose x component vector = As =A*Cosθ
Putting the value for given θ
As = 12.0 x Cos (90o +20o)
As = 120 x Sin(20o) = 12.0 * (-0.342) = -4.104 m
For the same way, component along y axis
Ay = 12.0 x Cos20o = 12.0 x 0.939 = 11.27 m
Similarly, for B vector component along the x and y axis
Bx = 20 *Cos(220o) = 20 * (-0.667) = -15.32 m
And By = 20 *Sin(220o) = 20 * (-0.6427) = -12.85 m
Therefore, resultant force in x and y axis
Rx = -15.32 –(-4.104) = -11.21 m
Ry = -11.27-12.85 = -24.12 m
The resultant vector
The resultant magnitude
R = √ ( 11.21 )2+ ( 24.12 )2 = √707.43=26.6 m Ans
And direction tanθ=−24.12
−11.21 , thenθ=65.1o Ans
Solution 8
Suppose there are three two dimensional vectors,
X =X x ^i+X y ^j+ X z ^k
Y =Y x ^i+Y y ^j +Y z ^k
Z=Z x ^i+Z y ^j+ Zz ^k
50 | P a g e

Numerical
Now Adding three vectors, like X+Y+Z
X+Y+Z = (X x ^i+ X y ^j+ Xz ^k ¿+¿(Y x ^i+Y y ^j +Y z ^k ¿+(Zx ^i+ Z y ^j+ Zz ^k ¿
And adding these vectors in different order
Z+Y+X = ( Zx ^i+ Z y ^j+Z z ^k ) +¿(Y x ^i+Y y ^j+Y z ^k ¿+ ( X x ^i+X y ^j+ Xz ^k ¿
= (X x ^i+ X y ^j+ Xz ^k ¿+¿(Y x ^i+Y y ^j+Y z ^k ¿+(Zx ^i+Z y ^j+Zz ^k ¿
In this condition we can say that, sum of these vector in different order results the same
Solution 9
The rectangular form of vector⃗
A=ACosθi+ ASinθj ⃗
A=27.5 cos 66o i +27.5 sin 66o j⃗
A=11.2i+25.1 j
Similarly⃗
B=BCosθi+ BSinθj⃗
B=30 cos 112o i +30 sin 112o j⃗
B=−11.24 i+27.81 j
Adding both vector
51 | P a g e

Numerical
R = ⃗ A+⃗ B
¿ ( 11.2i+25.1 j ) + ( −11.24 i+27.81 j ) =−0.04 i+52.9 j
The resultant magnitude
R= √ (−0.04 )2 + ( 52.9 )2= √2799.5=52.9m
The direction can be given as
tanθ= 52.09
−0.04 , thenθ=−89.9o
Angle w.r.t x direction
= 180-89.9o = 90.1o Ans
Solution 10
If we resolve the vector into x and y component, then vector in x direction can be given as
V total cos ( 22.5o +26.5o )=V A cos ( 22.5o )+V B cos ( 22.5o +26.5o+23o )
V total cos ( 49o )=V A cos (22.5o ) +V B cos (72o ) .........(i)
Similarly, in y direction
V total sin ( 49o )=V A sin ( 22.5o ) +V B sin ( 72o ) .......... (ii)
If we eliminate VB from equation (i) and (ii)
52 | P a g e

Numerical
V B = V total cos ( 49o ) −V A cos ( 22.5o )
cos ( 72o )
Similarly, for equation (ii),
V B = V total sin ( 49o ) −V A sin ( 22.5o )
sin (72o )
Equating the above two equation
[ V total sin ( 49o )−V A sin ( 22.5o ) ] sin ( 72o )= [ V total cos ( 49o )−V A cos ( 22.5o ) ] (cos 72o )
⟹−V A sin ( 72o−22.5o ) =−V total sin(72o −22.5o )
⟹ V A sin ( 49.5o ) =V total sin(23o)
Therefore, V A =V total sin ( 23o )
sin ( 49.5o ) Now putting the values
V A = 6.72sin ( 23o )
sin ( 49.5o ) =3.453 m/s Ans
Now putting the value of VA and VTotal
V B = 6.72cos ( 49o ) −3.453 cos (22.5o )
cos ( 72o ) =3.943 m/ s Ans
Solution 11
The component of VTotal can be given as
V Total ( x )=6.72 cos 49o=4.41m/s
V Total ( y ) =6.72 sin 49o=5.07 m/ s Ans
Solution 12
The resultant component of rotated axis can be given, the x component is given as
V Total ( x )=6.72 cos(22.5¿¿ o+26.5o−30o )=4.41m/ s ¿
53 | P a g e

Numerical
V Total ( x ) =6.72 cos(19o)=6.354 m/s Ans
And the y component
V Tot al ( y )=6.72sin (22.5¿¿ o+ 26.5o−30o)=5.07 m/s ¿
V Total ( y )=6.72 sin ( 19o )=2.188 m/ s
Solution 13a
As given in figure, 3.58
The distance traveled by C is given as
X = 1*120+5*120+2*120+1*120+1*120+3*120 = 1560 m = 1.56 km
Solution 13b
Resolving the component in x and y direction,
X s=0+5∗120+0−2∗120+0−3∗120=120 m
Y s =120+0−2∗120+0+120+0=0 m
Then resultant will
S= √ Xs
2+ X y
2 = √ 120
The angle can be given as
tanθ= 0
120 , thenθ=0o east Ans
Solution 14
The distance travelled along the path D
S = 2*120+6+120+4*120+1*120 = 1560 m
If we resolve the component into x and y direction
X s=0+720+ 0−120=600 m
54 | P a g e

Numerical
Y s =−240+0+4800=240m
S= √ Xs
2+ X y
2 = √ 6002 +2402=646.2m
And the direction tanθ= 240
600 , thenθ=21.8 north of east
Solution 15
Resolving the vectors in x and y components
X s=123∗sin 45o =86.97 km
X s=123∗cos 45o=86.97 km Ans
Solution 16
Resultant
55 | P a g e

Numerical
R= √ X2+ y2 Putting the values
R = √182 +252=30.8 m
The direction tanθ= 25
18 , thenθ=54.25o north of east The compass reading will be 90 –
54.25 = 35.75o west to north. Analytical technique is more accurate process.
Solution 17
As given in question,
R = X+Y = Y+X
The resultant R= √ X2+ y2= √ 252+ 182=30.8
And direction tanθ= 18
25 , thenθ=35.8o o westof north
In this condition we must take alternate path where there is not obstacle.
Solution 18a
As given in fig.
We can see that, <ABC = <BAD
Segment AC is in the direction east
The distance AC cos ( 90o−α ) = AC
AB
56 | P a g e

Numerical
AC= ABCos ( 90o−α ) ∨AB∗Sinα Putting the value in it
AC = 7.5 Sin15o = 1.94 km
Similarly CB = 7.5*Cos15o = 7.24 km
Solution 18a
In this condition the overall route must be ADB. In this
condition we can say that, the resulting displacement is still same as AB.
Solution 19a
As per figure the resultant displacement
R = X+Y = X-Y = 18i-25j
And its magnitude
R= √ X2+ y2= √ 252+ 182=30.8
57 | P a g e

Numerical
And direction tanθ= 25
18 , thenθ=54.2oo south of west
Solution 19b
As per figure the resultant displacement
R = X+Y = X-Y = -18i+25j
And its magnitude
R= √ X2+ y2= √ 252+ 182=30.8
And direction tanθ= 25
18 , thenθ=54.2o❑ south of west Ans
Solution 20
58 | P a g e

Numerical
As per the vector diagram,
The sum of vector = A+B+C = 0
Or C =-A-B
Resolving the vector into x and y component
X c=− ( ACos21o ) −BCos ( 101o ) =− ( 80∗cos21o ) −105∗cos ( 101o ) =−54.65 m
Y c=− ( ASin 21o ) −BSin ( 101o )=− ( 80∗sin21o )−105∗sin ( 101o ) =−74.40 m
The resultant vector C
And its magnitude
C= √ X2 + y2= √−54.652 +−74.402=92.3 m
And direction tanθ= −74.4
−54.65 , thenθ=53.7o❑ southof west Ans
Solution 21a
The component along south direction
59 | P a g e

Numerical
Ds = RSinθ = 32Sin(35o) = 18.4 km
DW = RCosθ = 32Cos(35o) = 26.2 km
Solution 21b
Similarly, distance towards South and north direction is SSW and SNW
SSW = RCosθ = 32Cos(10o) = 31.5 km
SNW = RSinθ = 32Cos(10o) = 5.56 km Ans
Solution 22a
As given in diagram, we
must resolve all the vector into x and y component
As per the vector diagram,
The sum of vector = A+B+C+D = 0
Or D =-A-B-C
Resolving the vector into x and y component
X D =− ( ACos−7.5o ) −BCos ( 90o +16o )−CCos ( 180o −19o ) =−¿
Y D=− ( ASin−7.5o )−BSin ( 90o+16o ) −CSin ( 180o−19o )=− ( 4.75 sin−7.5o ) −2.48 sin ( 106o ) −3.02 sin ( 161o )=−2.7
The resultant vector C
60 | P a g e

Numerical
And its magnitude
C= √ X2 + y2= √ −1.122 +−2.752=2.97 kmm
And direction tanθ=−1.12
−2.75 , thenθ=22.2o❑ swest of south Ans
Solution 23
The given vector can be Summerside as follows
i) 2.5 km at 135o
ii) 4.7 km at 300o
iii) 1.3 km at 205o
iv) 5.1 km at 0o
v) 1.7 km at 85o
vi) 7.2 km at 235o
vii) 2.8 km at 10o
The given problem can be represented as in fig
After resolving the given
61 | P a g e

Numerical
X R=2.5cos ( 135o ) + 4.7 cos ( 300o ) + 1.3cos ( 205o ) +5.1 ¿
Y R=2.5 sin ( 135o ) + 4.7 sin ( 300o )+ 1.3sin ( 205o ) +5.1 ¿
The resultant can be given as
C= √ X2 + y2= √ 3.282 +6.572 =7.34 km m
And direction tanθ=−6.57
3.28 ,then θ=−63.5o❑ south of east Ans
Worksheet Four – Kinematics
Solution 52a
As given in question, Suppose, displacement is S
Which can be given as S=vo t, Putting the value)
S = 3.53x 169*60 = 3.579 x 104 m Ans
Solution 52b
Suppose the velocity of person is XPG, velocity of person XPA and XAG is velocity of air
w.r.t ground
From given in question it can be written as
X PA= XPG −X AG Putting the values
X PA=3.53 — 2.0=5.53 m/s . and the direction is south east
Solution 52c
The displacement w.r.t air mass is given as
Y A =Y PA t=5.53∗169∗60=56.1 km Ans
Solution 53a
The different velocity w.r.t ground is given as
X SG=X SA +X AG
And X AG =X SG−X SA
62 | P a g e

Numerical
The velocity of seagull w.r.t ground
X AG = XSG
t =6.00 x 103
20∗60 =5.00 m
s
Then X AG =X SG−X SA=5.00−9.00=−4.00 m
s Ans
Solution 53b
X SG=X SA +X AG =−9.00−4.00=13.00 m/s
Time taken to return the seagull is
t= x SG
XSG
=6.00 x 103
13.00 =462 s=7 min 42 sec Ans
Solution 54a
Suppose the velocity of air 2nd runner is given as
X SF= XSG− XFG
Putting the value
X SF= XSG− XFG =4.2−3.5=0.70 m
s
Solution 54b
The time taken by 1st runner to finish the line
tf = xf
XFG
= 250
3.5 =71.43 sec
Similarly, time for 2nd runner
t'
f = x'
f
X FG
=250+ 45
4.2 =70.24 sec
As per result, the time taken by 2nd runner is less, therefore, 2nd runner will win.
63 | P a g e

Numerical
Solution 54c
The distance covered by second runner
X SF=−45+0.7 x 70.24=4.17 m ahead from first runner
Solution 55
The time taken to cover the distance 1.5 m by coin is given as
t= √−2 ( H −Ho )
g
And horizontal distance covered by coin
x−xo=v x t Putting the value of t and other value
x−xo=v x √−2 ( H−H o )
g =260 √−2 (−1.50 )
9.8 =144 m Ans
Solution 56
We know that, the range formula is given as
R= ( vg )2 sin ( 2θ )
g Putting the value from question
18= ( vg )2 sin ( 2θ )
9.8 ⟹ V b= √ 18∗9.8
sin 50o =15.2m/s Ans
Now as per vector given below
64 | P a g e

Numerical
The velocity of ball w.r.t player can be given as
V BA =V B−V A
V B =V BA + V A
The component of vectors cans be given as
V B cos θ=V BAx +V ax
V BAx=V B cos θ−V ax Putting the value
V BAx =15.2 cos ( 25o ) +2.00=15.8 m/ s
Now component in y axis,
V B sin θ=V BAy +V ay Putting the values
V BAy =V B sin θ−V ay=15.2 x sin θ ( 25o )−0=6.42m/s
Now resultant can be calculated as
vba= √ V BAx
2 +V BAy
2
vba= √ 15.82+6.422 =17.0 m/s
And direction tanθ= 6.42
15.8 , thenθ=22.1o
Ans
Solution 57
The vector representation of ship velocity is as follow
65 | P a g e

Numerical
XSW = XSG - XWG
The velocity of water relative to ground
XWG = 1.5Cos40oi+ 1.5Sin40oj
The velocity of ship w.r.t ground
XSG = XSW + XWG
The resolving the vectors in x and y component
XSGx = XSWx + XWGx
XSGy = XSWy + XWGy Now putting the values
XSGx = 0 +1.5Cos(40o) = 1.149 m/s
XSGy = 7 + 1.5*Sin(40o) = 7.964 m/s
Now the resultant of vector,
X SG= √ X SGx
2 + X SGy
❑ 2=8.05 m/ s
The direction of the vector =tanθ= 7.964
1.149 ,then tanθ=81.8o North of east
Solution 58
The velocity of jet w.r.t air is given as
XJA = XJG - XAG
Resolving the vectors in x and y axis.
XJax = 260*Cos(5o)i-260Sin(5o)j
66 | P a g e

Numerical
The velocity of air w.r.t ground
XAGx = 35*Cos(15o)i-35Sin(15o)j
And velocity of jet w.r.t ground
XJGx = -260Cos(5o)+35*Cos15o = -225.204 m/s
XJGy = -260Sin(5o)-35*Cos5o = 31.72 m/s
The resultant velocity will be
XJG = √ XJGx
2 + X JGy
2= √ 225.2042 +31.722=230 m/s
The direction will be
tanθ= 31.72
225.204 ,then θ=8.0o South of west
Solution 59a
As given in question,
The horizontal component of boat w.r.t ground
XS = 0 = XSX + XCCos40o = -= -XS.Sinφ+ XCCos40o
Now we can calculate, Sinφ = 1.5/7 Cos40o = 9.45o West of north
67 | P a g e

Numerical
Solution 59b
The velocity w.r.t earth is
YS = 0 = YSy + XCSin40o = 7.0 Cos(9.45o) +1.5 Sin(40o) = 7.87 m/sec Ans
Solution 60a
As given in figure,
After resolving the vector in x and y direction
V PGx=V PAx +V AGx
V PG∗cos (225¿¿ o)=V PA∗cos(185¿¿ o)+V AG∗cos(−200)¿ ¿ and
V PG∗sin (225¿¿ o)=V PA∗sin(185¿¿ o)+V AG∗sin(−200)¿ ¿
Multiply, Sin (225o) in x component and Cos (225o) in y direction
V PA∗sin(185¿¿ o)+V AG∗cos(−200) ¿ sin (225o)=¿
Putting the value VAG = 45 m/s
V PA∗sin( 185¿¿ o)+45∗cos (−200 )¿=¿
68 | P a g e

Numerical
V PA ¿
V PA∗sin ( 225o−185o ) =−45∗sin ( 245o ) VPA = 63.45 m/s Ans
Solution 60b
Asper the given case w must multiply Sin(185o) for horizontal component and Cos(185o)
to vertical component.
V PA∗¿
V PA= 45 sin(205o)
sin(40o) =29.59Ans
Solution 61a
As given in question,
V 2=u2+2 gh putting the value
v= √2 x 9.8 x 15=17.15m/ secAns
Solution 61b
As per question, the formula for sandal velocity
V sd =V sg−V dg
Now, sandal velocity w.r.t sandal
V sd =V sg +V dg
The magnitude for the above velocity
V sg = √ ¿ ¿ Now putting the value
V sg = √ ( 1.75 )2 + (−17.15 )2=17.2m/ s
And the direction is
tanθ= 17.15
1.75 , thenθ=84.2o South of east
69 | P a g e

Numerical
Solution 61c
The sandal and velocity of ship are same, that is why sandal looks like falling straight to
the ship, but for observer, the path of sandal is different due to different magnitude and direction.
Solution 62
Resolving the component in west direction
XWE = XWECos50o = 4.5Cos50o = -2.89 m/s
Similarly, for y direction
XWE = XWECos50o = 4.5Cos50o = -2.89 m/s
Now for Ocean velocity
XOE = XOESin30o = 2.2*Sin30o = 1.1 m/s
Component in north direction
XOE = XOECos30o = 2.2Cos50o = 1.905 m/s
The resolving the component of wind velocity w.r.t ocean in west east line, for x direction
XWO = XWE - XOE
XWOx = -2.89 – 1.1 = -3.99 m/s Similarly in y direction
XWOy = -3.44 – 1.9 = -5.34 m/s
The resultant of the velocity
70 | P a g e

Numerical
XWO = √ ( XWOx )
2 + ( XWOy )
2= √ ( −3.99 ) 2 + ( −5.34 ) 2=6.67 m/s
And the direction is
Tanθ= 5.352
3.993 , thenθ=53.3o South of west Ans
Solution 63a
The relative velocity of galaxy 1 is calculated as
A12 = A1 – A2 = -4500+2200 = -2300 km/s
The relative velocity of galaxy 2 is calculated as
A22 = A2 – A2 = 0 km/s
The relative velocity of galaxy 3 is calculated as
A32 = A3 – A2 = 0-2200 km/s
The relative velocity of galaxy 4 is calculated as
A42 = A4 – A2 = 2830 +2200 = 5030 km/s
The relative velocity of galaxy 5 is calculated as
A52 = A5 – A2 = 6700 +2200 = 8900 km/s Ans
Solution 63b
The relative velocity of galaxy 1 w.r.t Galaxy 5 is calculated as
A15 = A1 – A5 = -4500-6700 = -11200 km/s
The relative velocity of galaxy 2 w.r.t Galaxy 5 is calculated as
A25 = A2 – A5 = -2200-6700 = -8900 km/s
The relative velocity of galaxy 3 w.r.t Galaxy 5 is calculated as
A35 = A3 – A5 = -0-6700 = -6700 km/s
The relative velocity of galaxy 4 w.r.t Galaxy 5 is calculated as
A45 = A4 – A5 = 2830-6700 = -3870 km/s
71 | P a g e

Numerical
The relative velocity of galaxy 4 w.r.t Galaxy 5 is calculated as
A55 = A5 – A5 = 0-0 = 0 km/s Ans
Solution 64a
The formula for rate of expansion = H= v
d
Now for Galaxy 1, putting the value
Hubble Constant for Galaxy 1
H1= −4500
−300 x 106 x 9.46 x 1012 =1.59 x 10−18 / s
Hubble Constant for Galaxy 2
H2= −−2200
−150 x 106 x 1 9.46 x 1012 =1.55 x 10−18 / s
Hubble Constant for Galaxy 4
H4 = 2830
190 x 106 x 9.46 x 1012 =1.57 x 10−18/ s
Hubble Constant for Galaxy 5
H5= 6770
450 x 106 x 9.46 x 1012 =1.59 x 10−18 / s
The average Hubble constant
Haverage =1.59 x 10−18+1.55 x 10−18+ 1.57 x 10−18 +1.59 x 10−18
4 =1.57 x 10−18 /s
Solution 64b
The time can be calculated by inverse of H
t= 1
1.57 x 10−18 x 6.3 7 x 1017
3.16 x 107 =2.02 x 1010 years Ans
72 | P a g e

Numerical
Solution 65a
As given in question,
Time taken by swimmer = 25/0.5 = 50 sec
In this condition speed of river current
VWG = 40/50 = 0.8 m/s
Vector addition of velocities resultant of magnitude
VSG = √0.52 +0.82=0.943 m/ s
Solution 66
The vector diagram of given problem can be represented as follows
The horizontal and vertical component of water velocity w.r.t earth
VWEx = -VSWCos115o + -VWECos95o = 4.0*Cos115o+4.8Cos95o = 1.272 m/s
VWEx = -VSWSin115o + -VWESin95o = -4.0*Sin115o+4.8Sin95o = 1.157 m/s
The resultant magnitude of water velocity
VWE = √ 1.2722+1.1572=1.72 m/s
And direction,tanα= 1.157
1.272 ,then α=42.3o North of east
Solution 67
The vector of the given problem is as follows
73 | P a g e

Numerical
From above figure, we can calculate
VPU Sinθ= VPL
In this condition θ = Sin-1(8/29) = 16.01o Ans
The angle β = 90o -θ = 74.0o Ans
Solution 68a
The launch velocity can be given by
1
2 m v2=GMm ( 1
R − 1
h+R )
v2=2GM ( h+R−R
( R ) ( h+R ) )=2 GM ( h
( R ) ( h+R ) )
v= √2GM ( h
( R ) ( h+ R ) )Now putting the value in Formula.
v=
√2(6.67 x 10−11)(5.972 x 1024) ( 36 x 106
( 6.37 x 106 ) ( 36 x 106 +6.37 x 106 ) ) = 10.31 km/s Ans
Solution 68b
Since, the escape velocity on the earth surface 11.174 km/s, and the calculated velocity is 10.31,
this will never reach beyond earth.
Solution 68c
As the height increases, the magnitude is decreases
74 | P a g e

Numerical
Solution 68d
The given assumption is not valid due to height beyond gravitational pull, mgh will work
because for astronomy, increase in height decreases forces.
Solution 69a
The velocity of place w.r.t ground is calculated as
vpg =3000 x 103
1.5 x 3600 =556 m/ s
Solution 69b
The x component of air velocity w.r.t ground
V AG , x=556∗cos ( −5o ) −280cos ( oo ) =273.44 m/ s
Similarly, y component
V AG , y=556∗sin ( −5o ) −280 sin ( oo ) =−48.42 m/ s
Now magnitude,
V AG , y= √ ( V AG , x )2+ ( V AG , y ) 2= √273.442 + (−48.42 )2 =277.7 m/s
And direction,tanθ=−48.42
273.44 ,then θ=10.04o South east
Solution 69c
The magnitude can be calculated from total velocity and wind velocity, are unreasonably
very high
Solution 69d
The distance is 3000 km, and time is 1.5 h, it means the velocity should be very large, which is
also unreasonable.
75 | P a g e